How does `cout` decide whether to print the address or the string when you pass a character pointer?

Understanding Stream Operator Overloading for Char Pointers

When printing a character pointer using printf(), the conversion specifier determines whether the address or the string is printed, such as %u for the address or %s for the string. However, with C streams and cout, how does it decide which one to output?

Consider the following code:

<code class="cpp">char ch = 'a';
char *cptr = &ch;
cout <p>In this example, cout attempts to interpret cptr as a string. To print the address of ch using cout, a conversion must be applied to resolve this ambiguity.</p>
<p><strong>Solution: Casting to a Void Pointer</strong></p>
<p>The preferred method for obtaining an address using cout is through type casting. The correct approach is:</p>
<pre class="brush:php;toolbar:false"><code class="cpp">cout (cptr) <p>By casting cptr to void *, we force cout to execute the appropriate overload that takes a void pointer (ostream& operator</p></code>

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